$g^x = g^{x\bmod m}$ in a finite group of order $m$

Question

I want to understand the proof that $g^x = g^{x\bmod m}$ for $g\in G$ for $G$ such that $\vert G\vert = m$

The proof goes as follows

say that $x = qm+r$

then $$g^x = g^{qm+r} = g^{qm}g^r = (g^m)^qg^r = 1^qg^r = g^r = g^{x\bmod m}$$

How is $g^m = 1$? Or is there an hypothesis missing? Such as $g$ is a generator.

Moreover, if I take $\mathbb{Z}^*_5$, then $2^5 = 2 \ne 1 = 2^{5\bmod 5}$

Answer 1

In a finite group $G$ of order $m$ we have $g^m=e$ for all $g\in G$. This follows from Lagrange's Theorem. A special case is for $G=(\mathbb{Z}/n\mathbb{Z})^{\times}$, of order $m=\phi(n)$, which is called Euler's Theorem.

Answer 2

The order of the group generated by $g$, $<g>$, must divide the order of the group, by Lagrange's theorem. But the order of $<g>$ is the order of $g$. Set $d=\lvert <g >\rvert $. Thus $m=k\cdot d$ for $k\in \mathbb Z$... So $g^m=g^{kd}=(g^d)^k=(e)^k=e$...

Secondly, $\phi (5)= 5-1=4$.